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True or False Every polynomial with real numbers as coefficients can be factored into products of linear and/or irreducible quadratic factors.

Short Answer

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True.

Step by step solution

01

Understand Polynomial Factorization

Polynomials with real coefficients can sometimes be factored into linear polynomial factors (degree 1) and quadratic polynomial factors (degree 2). Understanding what it means to have 'irreducible quadratic factors' is crucial; it means those quadratic polynomials cannot be factored further over the real numbers.
02

The Fundamental Theorem of Algebra

According to the Fundamental Theorem of Algebra, every non-constant polynomial with real coefficients can be factored into a product of linear factors, possibly involving complex numbers if not all roots are real.
03

Factorization Over Real Numbers

When we restrict our coefficients to real numbers, some polynomials do not have all real roots and thus cannot be fully factored into linear factors with real coefficients. Instead, some factors will be irreducible quadratic factors.
04

Conclusion

Thus, every polynomial with real coefficients can indeed be factored into a product of linear factors and/or irreducible quadratic factors over the real numbers. This means the given statement is true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Irreducible Quadratic Factors
When working with polynomial factorization, an important concept to grasp is 'irreducible quadratic factors'. These are quadratic polynomials that cannot be factored further over the real numbers.

For example, the quadratic polynomial \( x^2 + 1 \) is irreducible over the real numbers because it has no real roots.
This means it cannot be expressed as a product of two linear factors with real coefficients.

In general, an irreducible quadratic factor in the context of real numbers is a polynomial of the form \( ax^2 + bx + c \) (where \(a, b,\) and \(c\) are real numbers), which does not have real roots. A quick way to check if a quadratic polynomial is irreducible is to use the discriminant.

The discriminant of a quadratic polynomial \( ax^2 + bx + c \) is given by:
\[ D = b^2 - 4ac \]

If \( D < 0 \), the quadratic polynomial does not have real roots, thus making it irreducible over the real numbers.

This concept is crucial in proving that any polynomial with real coefficients can be factored into linear and/or irreducible quadratic factors.
Fundamental Theorem of Algebra
The Fundamental Theorem of Algebra is foundational in understanding polynomial factorization. This theorem states that every non-constant polynomial with complex coefficients has at least one complex root.

In simpler terms, it guarantees that any polynomial equation of degree \(n\) has \(n\) roots in the complex number system (counting multiplicity).

For instance, a polynomial of degree 3 will always have exactly 3 roots, though some might be repeated or non-real.

When dealing with real coefficients, if a polynomial has complex roots, they will occur in conjugate pairs due to the complex conjugate root theorem. For example, if \( a + bi \) is a root, \( a - bi \) will also be a root.

Because of this theorem, we understand that all polynomials can be factored into a product of linear factors (if roots are real) and/or quadratic factors (if roots are complex conjugates) over the real numbers.

Thus, even though the theorem's primary focus is on complex coefficients and roots, it indirectly supports the understanding that polynomials with real coefficients can be expressed using linear and irreducible quadratic factors.
Real Coefficients
When a polynomial has real coefficients, the nature of its factorization changes compared to polynomials with complex coefficients.

Real coefficients mean that each term in the polynomial is a real number. This affects how we factorize the polynomial when focusing solely on real numbers.

If we restrict our factorization to real coefficients, we may encounter irreducible quadratic factors. These are quadratic polynomials that don't have real roots.

Suppose we have a polynomial like \( x^4 + 4 \). While it doesn't factor into polynomials with real coefficients easily, it can be broken down into two irreducible quadratics:
\[ x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2) \]

Each of these quadratic factors is irreducible over the real numbers because their discriminants are negative.

In summary, polynomials with real coefficients can be fully factorized into linear and irreducible quadratic factors. This conclusion complements our understanding of the Fundamental Theorem of Algebra and polynomial factorization.

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Most popular questions from this chapter

Three retired couples each require an additional annual income of \(\$ 2000\) per year. As their financial consultant, you recommend that they invest some money in Treasury bills that yield \(7 \%\), some money in corporate bonds that yield \(9 \%,\) and some money in "junk bonds" that yield \(11 \%\). Prepare a table for each couple showing the various ways that their goals can be achieved: (a) If the first couple has \(\$ 20,000\) to invest. (b) If the second couple has \(\$ 25,000\) to invest. (c) If the third couple has \(\$ 30,000\) to invest. (d) What advice would you give each couple regarding the amount to invest and the choices available?

Carletta has \(\$ 10,000\) to invest. As her financial consultant, you recommend that she invest in Treasury bills that yield \(6 \%,\) Treasury bonds that yield \(7 \%\) and corporate bonds that yield \(8 \% .\) Carletta wants to have an annual income of \(\$ 680\), and the amount invested in corporate bonds must be half that invested in Treasury bills. Find the amount in each investment.

Verify that the values of the variables listed are solutions of the system of equations. $$ \begin{array}{l} \left\\{\begin{array}{l} 4 x -5 z=6 \\ 5 y-z =-17 \\ -x-6 y+5 z =24 \\ \end{array}\right.\\\ x=4, y=-3, z =2 ;(4,-3,2) \end{array} $$

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{r} 2 x-3 y-z=0 \\ -x+2 y+z=5 \\ 3 x-4 y-z=1 \end{array}\right. $$

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{l} 3 x-2 y+2 z=6 \\ 7 x-3 y+2 z=-1 \\ 2 x-3 y+4 z=0 \end{array}\right. $$

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